Question

The complete solution set of the inequality ${\log }_3(x+2)(x+4)+{\log }_{1/3}(x+2)<\frac{1}{2}{\log }_{\sqrt{3}}7$ is equal to

• A. $(-2-1)$
• B. $(-2,3)$
• C. $(-1,3)$
• D. $(3,\infty )$

Answer 1

The correct option is B $(-2,3)$

${\log }_3(x+2)(x+4)+{\log }_{1/3}(x+2)<\left(1/2\right){\log }_{\sqrt{3}}7$
Clearly, this is defined for $x>-2$
$\frac{\log (x+2)}{\log 3}+\frac{\log (x+4)}{\log 3}+\frac{\log (x+2)}{\log \left(\frac{1}{3}\right)}<\frac{1}{2}\frac{\log 7}{\frac{1}{2}\log 3}$
$\frac{\log (x+2)}{\log 3}+\frac{\log (x+4)}{\log 3}-\frac{\log (x+2)}{\log 3}<\frac{\log 7}{\log 3}$
$\Rightarrow \frac{\log (x+4)}{\log 3}<\frac{\log 7}{\log 3}$
$\Rightarrow {\log }_3(x+4)<{\log }_3\left(7\right)$
$x+4<7$
$\Rightarrow x<3$
The solution set is $(-2,3)$